ap calculus ab chapter 4: applications of derivatives section 4.2: mean value theorem
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AP CALCULUS AB
Chapter 4:Applications of Derivatives
Section 4.2:Mean Value Theorem
What you’ll learn about Mean Value Theorem Physical Interpretation Increasing and Decreasing Functions Other Consequences
…and whyThe Mean Value Theorem is an important
theoretical tool to connect the average and instantaneous rates of change.
Mean Value Theorem for Derivatives
If ( ) is continuous at every point of the closed interval , and
differentiable at every point of its interior , , then there is at least
( ) - ( )one point in , at which '( ) .
-
y f x a b
a b
f b f ac a b f c
b a
This means…….IFy = f(x) is continuousy = f(x) is on a closed interval [a,b]y = f(x) is differentiable at every point in its
interior (a,b)
THENSomewhere between points A and B on a
differentiable curve, there is at least one tangent line parallel to chord AB.
Example 1: Exploring the Mean Value TheoremShow that the function f(x) = x2 satisfies the hypothesis of
the Mean Value Theorem on the interval [0,2]. Then find a solution c to the equation
Consider f(x) = x2. Is it continuous? Closed interval? Differentiable?
If so, by the MVT we are guaranteed a point c in the interval [0,2] for which ab
afbfcf
)()(
)('
ab
afbfcf
)()(
)('
Example 1 continuedGiven f(x) = x2
Interval [0,2]
To use the MVT Find the slope of the
chord with endpoints (0, f(0)) and (2, f(2)).
Find f’
Set f’ equal to the slope of the chord, solve for c
Find c
ab
afbfcf
)()(
)('
Interpret your answer The slope of the tangent line to f(x) = x2 at x = 1 is equal to the slope of the chord AB. OR The tangent line at x = 1 is parallel to chord AB.
Write equations for line AB and the tangent line of y = x2 at x=1.
Graph and investigate. The lines should be parallel.
Example Explore the Mean Value Theorem
2Show that the function ( ) satisfies the hypothesis of the Mean Value
Theorem on the interval 0,2 . Then find a solution to the equation
( ) - ( )'( ) on this interval.
-
f x x
c
f b f af c
b a
2The function ( ) is continuous on 0,2
and differentiable on 0,2 .
Since (0) 0, (2) 4, and '( ) 2
( ) - ( )'( )
-4 0
22 0
2 2
1.
f x x
f f f x x
f b f af c
b a
c
c
c
Example 2: Mean Value Theorem Ex:
.12
5 So
2
5
52
142
122
114
14
4244
1211 22
4,1on 2
2
2
2
fc
c
c
mc
m
f
fxxf
xxf
Rolle’s Theorem:If f is continuous on the closed intervaland differentiable on the open intervaland if then there is at least one number c insuch that
Rolle’s Theorem
ba, ba,
bfaf ba,
.0 cf
a b
(a, f(a))
(b, f(b)
Show that the function satisfies the hypothesis of the MVT on the interval [0,1]. Then find c
1. Is it continuous? Closed interval? Differentiable?
2. Find c
3. Interpret your findings
12)( 2 xxxf
Example 2: Further Exploration of the MVTExplain why each of the following functions
fails to satisfy the conditions of the Mean Value Theorem on the interval [-1,1].
You try:
1)( 2 xxf
1,1
1,3)(
2
3
xx
xxxf
3
1
)( xxf
Example 3: Applying the Mean Value Theorem
Find a tangent to f in the interval (-1,1) that is parallel to the secant AB.
Given: , A = (-1,f(-1)) and B = (1, f(1))
1) Find the slope of AB and f’(c)2) Apply MVT to find c3) Evaluate f(x) at c, use that point and the slope
from step 1 to find the equation of the tangent line
21)( xxf
You Try - 1) Find slope of chord AB that connects endpoints
2) Find f ’ and apply MVT formula to find c.
3) Evaluate f(x) at c, use that point and the slope from step 1 to find the equation of the tangent line. Graph & check.
31 x1)( xxf
Physical Interpretation of the Mean Value Theorem
The MVT says the instantaneous change at some interior point must equal the average change over the entire interval.
f’(x) = instantaneous change at a point = average change over the
interval
ab
afbf
)()(
Example 4: Interpreting the MVTIf a car accelerating from zero takes 8 sec to
go 352 ft, its average velocity for the 8-second interval is 352 / 8 = 44 ft/sec, or 30 mph.
Can we cite the driver for speeding if he / she is in a residential area with a speed limit of 25 mph?
hour
miles
hourft
mileft 30
1
sec3600
5280
1
sec
44
Increasing Function, Decreasing Function
1 2
1 2 1 2
1 2 1 2
Let be a function defined on an interval and let and
be any two points in .
1. on if ( ) ( ).
2. on if ( ) ( ).
f I x x
I
f I x x f x f x
f I x x f x f x
increases
decreases
Corollary: Increasing and Decreasing Functions
Let be continuous on , and differentiable on , .
1. If ' 0 at each point of , , then increases on , .
2. If ' 0 at each point of , , then decreases on , .
f a b a b
f a b f a b
f a b f a b
Example Determining Where Graphs Rise or Fall
3Where is the function ( ) 2 12 increasing and where is it decreasing?f x x x
2
2
2
2
The function is increasing when '( ) 0.
6 12 0
2
2 or 2
The function is decreasing when '( ) 0.
6 12 0
2
2 2
f x
x
x
x x
f x
x
x
x
Corollary: Functions with f’=0 are Constant If '( ) 0 at each point of an interval , then there is a
constant for which ( ) for all in .
f x I
C f x C x I
Section 4.2 – Mean Value Theorem To find the intervals on which a function is
increasing or decreasing1. Locate the critical numbers of f in Use
these numbers to determine the test intervals. (when or is undefined, and the endpoints of the interval).
2. Determine the sign of at a value in each of the test intervals.
3. Decide whether f is increasing or decreasing on each interval.
.,ba
0 xf
xf
f’(x) a + b - c
f(x) inc. dec.
Section 4.2 – Mean Value Theorem Ex:
1 ,3
1 :numbers Critical
01 013
0113
113
143
522
23
x
xx
xx
xx
xxxf
xxxxf
1/3 1f’(x)
f(x)
+ +
inc. dec. inc.
.1,3
1on decreasing is
.,13
1,on increasing is
xf
xf
Example 5 Determining where graphs rise or fall
Use corollary 1 to determine where the graph of f(x) = x2 – 3x is increasing and decreasing.
1) Find f’ and set it equal to zero to find critical points.
2) Where f ’ > 0, f is increasing.3) Where f ’ < 0, f is decreasing.4) Any maximum or minimum values?
Example 6 Determining where graphs rise or fall
Where is the function increasing and where is it decreasing?
Graphically: Use window [-5,5] by [-5,5]Confirm Analytically: Find f ’, evaluate f ’ = 0 to find critical points.
Where f ’ > 0, f is increasing. Where f ’ < 0, f is decreasing.
xxxf 4)( 3
Corollary: Functions with the Same Derivative Differ by a Constant If '( ) '( ) at each point of an interval , then there is a constant
such that ( ) ( ) for all in .
f x g x I
C f x g x C x I
. then ,3 if So,
31 35
32
2323
Cxyxdx
dy
xxdx
dxx
dx
d
Example:
Antiderivative A function ( ) is an of a function ( ) if '( ) ( )
for all in the domain of . The process of finding an antiderivative
is .
F x f x F x f x
x f
antiderivative
antidifferentiation
Example 7 Applying Corollary 3
Find the function f(x) whose derivative is sin x and whose graph passes through the point (0,2).
Write f(x) = antiderivative function + C Use (x,y) = (0, 2) in equation, solve for C.Write f(x), the antiderivative of sin x through (0, 2)
Find the Antiderivative!Find the antiderivative of f(x) = 6x2.
Reverse the power rule Add 1 to the exponent Divide the coefficient by
(exponent + 1) Add C You have found the
antiderivative, F(x)
Find the antiderivative of f(x) = cos x
Think Backwards: What function has a derivative of cos x?
Add C
You have found the antiderivative, F(x).
Find each antiderivative – don’t forget C!
1) f’(x) = -sin x
2) f’(x) = 2x + 6
3) f’(x) = 2x2 + 4x - 3
Specific AntiderivativesFind the antiderivative ofthrough the point (0, 1).
Write equation for antiderivative + c Insert point (0,1) for (x, y) and solve for c Write specific antiderivative.
3
1
3
2)( xxf
Example Finding Velocity and Position
2
Find the velocity and position functions of a freely falling body for
the following set of conditions:
The acceleration is 9.8 m/sec and the body falls from rest.
Assume that the body is released at time 0.
Velocity: We know that ( ) 9.8 and (0) 0, so
( ) 9.8
(0) 0
0. The velocity function is ( ) 9.8 .
Position: We know that ( ) 9.8 and (0) 0,
t
a t v
v t t C
v C
C v t t
v t t s
2
2
so
( ) 4.9
(0) 0
0. The position function is ( ) 4.9 .
s t t C
s C
C s t t
The acceleration is 9.8 m/sec2 and the body is propelled downward with an initial velocity of 1 m/sec2.
Work backwards
Velocity function + c, P(0,1) (why?)
Position function + c (from velocity function)
SummaryThe Mean Value Theorem tells us that If y = f(x) is continuous at every point on the closed interval [a, b] and differentiable at
every point of its interior (a,b), then there is at least one point c in (a,b) at which(The derivative at point c = the slope of the chord.)
It’s corollaries go on to tell us that where f ‘ is positive, f is increasing, where f ‘ = 0, f is a constant function, and where f ‘ is negative, f is decreasing.
We can work backwards from a derivative function to the original function, a process called antidifferentiation. However, as the derivative of any constant = 0, we need to know a point of the original function to get its specific antiderivative. Without a point of the function all we can determine is a general formula for f(x) + C.